For the Heaviside function: e^(-cs)*F(s)=u(t-c)*f(t-c) I have 2 questions.
- How is f(t-c) evaluated? If for instance my f(s) is 1/(s-2) and my e^(-cs) was e^(-3s) the answer should be u(t-3)*e^(2(t-3)). Apart from seeing that constant of 2 being left outside and a 3 being subtracted from the "t" I'm not getting what is happening. Is it f(t) - { f(t) evaluated at c}?
I'm reading that there is a shift of (3 units) happening here but that does not help me understand.
2. Is that value of C from the equation always positive?
Thank You.
The inverse Laplace transform of $$e^{-cs}F(s)$$ is $$ u(t-c)f(t-c)$$ Thus$$\frac {e^{-3s}}{s-2}$$ will change to $$ u(t-3)e^{2(t-3)}$$
The answer to your second question is yes, we shift functions to the right so c is positive.